Question

If $2^k$ > 111111.11111 , then the minimum integral value of k that satisfies the given condition is (given that log 2 = 0.3010 and log 3 = 0.4771 )

• A. 17
• B. 16
• C. 18
• D. 19

Answer 1

The correct option is A

17

$2^k$ > 111111.11111

log $2^k$ > log 111111.11111

k log2 > log $\left(\frac{1000000}{9}\right)$

0.3010 k > log${10}^6$ - 2log3

$0.3010k>6-2\times$ 0.4771

k > 16.76

So, minimum integral value of k that satisfies the relation is 17.